Answer :

Given: 2 (a2 + b2)x2 + 2(a + b)x + 1 = 0

Comparing with standard quadratic equation ax2 + bx + c = 0


a = 2 (a2 + b2), b = 2(a + b), c = 1


Discriminant D = b2 – 4ac


= [2(a + b)]2 – 4. 2 (a2 + b2).1


= 4(a2 + b2 + 2ab) – 8 a2 – 8b2


= 4a2 + 4b2 + 8ab – 8a2 – 8b2


= – 4a2 – 4b2 + 8ab


= – 4(a2 + b2 – 2ab)


= – 4(a – b)2 < 0


Hence the equation has no real roots.


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